y^{''}+4y^'+6y=cos(x),y(0)=0,y^'(0)=2

Lösung

y^{^{''}} + 4 y^{^{'}} + 6 y = cos (x), y (0) = 0, y^{^{'}} (0) = 2

y = e^{- 2 x} (- \frac{5}{41} cos (2 x) + \frac{34 2}{41} sin (2 x)) + \frac{4}{41} sin (x) + \frac{5}{41} cos (x)

Schritte zur Lösung

y^{''} (x) + 4 y^{'} (x) + 6 y = cos (x)

Solve linear ODE:

y = e^{- 2 x} (- \frac{5}{41} cos (2 x) + \frac{34 2}{41} sin (2 x)) + \frac{4}{41} sin (x) + \frac{5}{41} cos (x)

y = e^{- 2 x} (- \frac{5}{41} cos (2 x) + \frac{34 2}{41} sin (2 x)) + \frac{4}{41} sin (x) + \frac{5}{41} cos (x)

n = 1 \sum \infty \frac{x ^{n}}{2 n - 1}

\frac{\partial}{\partial x} (e^{x z})

x \to 0 + lim ((1 - 4 x)^{\frac{1}{x}})

\int 6 x (3 x^{2} + 4)^{4} d x

\frac{d}{d x} (\frac{( x - 2 ) ^{3}}{x ^{2}})